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The conductor of a cyclic quartic field using Gauss sums
Authors:Blair K Spearman  Kenneth S Williams
Institution:(1) Department of Mathematics and Statistics, Okanagan University College, Kelowna, B.C, Canada, V1V 1V7;(2) Department of Mathematics and Statistics, Carleton University, Ottawa, Ontario, Canada, K1S 5B6
Abstract:Let Q denote the field of rational numbers. Let K be a cyclic quartic extension of Q. It is known that there are unique integers A, B, C, D such that 
$$K = Q\left( {\sqrt {A(D + B\sqrt D )} } \right)$$
where A is squarefree and odd, D=B 2+C 2 is squarefree, B 
$$ > $$
0 , C 
$$ > $$
0, GCD(A,D)=1. The conductor f(K) of K is f(K) = 2 l |A|D, where 
$$l = \left\{ \begin{gathered}  3,{\text{   if }}D \equiv 2{\text{ }}({\text{mod 4}}){\text{ or }}D \equiv 1{\text{ (mod 4), }}B \equiv 1{\text{ }}({\text{mod 2}}), \hfill \\  2,{\text{   if }}D \equiv 1{\text{ (mod 4), }}B \equiv 0{\text{ (mod 2), }}A + B \equiv 3{\text{ (mod 4),}} \hfill \\  0,{\text{   if }}D \equiv 1{\text{ (mod 4), }}B \equiv 0{\text{ (mod 2), }}A + B \equiv 1{\text{ (mod 4)}}{\text{.}} \hfill \\ \end{gathered}  \right.$$
A simple proof of this formula for f(K) is given, which uses the basic properties of quartic Gauss sums.
Keywords:
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